Question

创建表单以通过HTML表单更新特定行数据的部分工作。我已成功从mysql数据库中检索行数据但是当我尝试在html表单的字段中显示特定的行数据（StudentNumber，FirstName，LastName等）时，它只显示空字段。我可能错过了什么？

以下是显示表单的php页面的编码：

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" 
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> 
<html xmlns="http://www.w3.org/1999/xhtml"> 
<head> 
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> 
<title>updating data record</title>
</head> 
<body>

<?php
//connect to the database
  $conn = mysql_connect('localhost', 'root', '');
    if (!$conn) {
    die('Not connected : ' . mysql_error());
}   
    $db_selected = mysql_select_db('ccm3413', $conn);
    mysql_set_charset("utf8",$conn);
    $id = $_GET['id'];

    $result = mysql_query("SELECT * FROM studentinfo WHERE recordID = '$id'");
    while ($row = mysql_fetch_assoc($result)) {
    }

?>

<form id="form1" name="form1" method="post" action="processUpdateRecord.php"> 
  <p> 
    <label>Student Number: 
      <input name="StudentNumber" type="text" id="StudentNumber" size="10" value="<?php echo $row['StudentNumber']; ?>"/> 
    </label> 
  </p> 
  <p> 
    <label>First Name: 
      <input name="FirstName" type="text" id="FirstName" size="20" value="<?php echo $row['FirstName']; ?>"/> 
    </label> 
  </p> 
  <p> 
    <label>Last Name: 
      <input name="LastName" type="text" id="LastName" size="20" value="<?php echo $row['LastName']; ?>"/> 
    </label> 
  </p> 
  <p> 
    <label>Email Address: 
      <input name="EmailAddr" type="text" id="EmailAddr" size="50" value="<?php echo $row['EmailAddr']; ?>"/> 
    </label> 
  </p> 
  <p> 
    <label>Telephone: 
      <input name="PhoneNumber" type="text" id="PhoneNumber" size="20" value="<?php echo $row['PhoneNumber']; ?>"/> 
    </label> 
  </p> 
  <p> 
    <input type="submit" name="Update" id="Update" value="Update" /> 
  </p>

</form>

</body>
</html>

Answer 1

请尝试使用此代码：

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" 
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> 
<html xmlns="http://www.w3.org/1999/xhtml"> 
<head> 
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> 
<title>updating data record</title>
</head> 
<body>

<?php
//connect to the database
  $conn = mysql_connect('localhost', 'root', '');
    if (!$conn) {
    die('Not connected : ' . mysql_error());
}   
    $db_selected = mysql_select_db('ccm3413', $conn);
    mysql_set_charset("utf8",$conn);
    $id = mysql_real_escape_string($_GET['id']);

    $result = mysql_query("SELECT * FROM studentinfo WHERE recordID = '$id'");
    $student = (mysql_num_rows($result)==1) ? mysql_fetch_assoc($result) : null ; //You expect only 1 student, right?

if (is_array($student)){
?>

<form id="form1" name="form1" method="post" action="processUpdateRecord.php"> 
  <p> 
    <label>Student Number: 
      <input name="StudentNumber" type="text" id="StudentNumber" size="10" value="<?php echo $student['StudentNumber']; ?>"/> 
    </label> 
  </p> 
  <p> 
    <label>First Name: 
      <input name="FirstName" type="text" id="FirstName" size="20" value="<?php echo $student['FirstName']; ?>"/> 
    </label> 
  </p> 
  <p> 
    <label>Last Name: 
      <input name="LastName" type="text" id="LastName" size="20" value="<?php echo $student['LastName']; ?>"/> 
    </label> 
  </p> 
  <p> 
    <label>Email Address: 
      <input name="EmailAddr" type="text" id="EmailAddr" size="50" value="<?php echo $student['EmailAddr']; ?>"/> 
    </label> 
  </p> 
  <p> 
    <label>Telephone: 
      <input name="PhoneNumber" type="text" id="PhoneNumber" size="20" value="<?php echo $student['PhoneNumber']; ?>"/> 
    </label> 
  </p> 
  <p> 
    <input type="submit" name="Update" id="Update" value="Update" /> 
  </p>
</form>
<?php

else {
  echo "The student has not been found" ;
} ?>

</body>
</html>

Answer 2

您需要将表单放在while括号内。

MYSQL行数据不会在HTML表单字段中回显/显示

2 个答案: